工厂的增量，拉拉威尔

Question

我有一个工厂被调用了好几次，但在contract_year列上我得到了相同的值。

$factory->define(ContractYear::class, function (Faker $faker) {
$contract = Contract::orderByDesc('id')->first();
$contract_year = ContractYear::select('contract_year')->orderByDesc('id')->value('contract_year');
if($contract_year == null){
$contract_year = 2019;
}
return [
'contract_id'                   => $contract->id,
'contract_year'                 => $contract_year++,
'licensed_users'                => $faker->randomDigit,
];
});

我从这里开始称呼它。。

$u->contracts()->saveMany(factory(Contract::class, rand(1, 5))->create()->each(function ($contract){
$contract->years()->saveMany(factory(ContractYear::class, $contract->number_of_years)->create());            
}));

Answer 1

您可以使用static

$factory->define(ContractYear::class, function (Faker $faker) {
static $contract_year;

$contract = Contract::orderByDesc('id')->first();
if(!$contract_year) {
$contract_year =  ContractYear::select('contract_year')->orderByDesc('id')->value('contract_year') ?? 2019;
}
return [
'contract_id'                   => $contract->id,
'contract_year'                 => $contract_year++,
'licensed_users'                => $faker->randomDigit,
];
});

引用：函数中的"static"关键字？

Answer 2

将计数器连接到$this

$factory->define(ContractYear::class, function (Faker $faker) {
$this->sequences = $this->sequences ?? [];
$contract = Contract::orderByDesc('id')->first();
$group = $contract->id ?? 0;
$this->sequences[$group] = $this->sequences[$group] ?? 2019;
return [
'contract_id'                   => $contract->id,
'contract_year'                 => $this->sequences[$group]++,
'licensed_users'                => $faker->randomDigit,
];
});

可能值得一看Laravel 8新的模型工厂，以获得更明确的面向对象方法